RC Transfer function

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alexmath

Joined May 2, 2014
17
Suppose R = C = 1 then the transfer function from the input voltage to the voltage across the capacitor is 1/ (s+1). So Vc(S) / V(S) = 1/(s+1). Getting back to time-domain: Vc(t) = V(t) * e^-t. What's wrong here? Thank you!
 
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MrAl

Joined Jun 17, 2014
11,472
Hello there,

Constant voltage sources V(s) transform into V/s they do not stay the same when there is a step input. So V(s) => V/s so your Laplace is not V/(s+1) it is (V/s)/(s+1)=V/(s*(s+1)). If the voltage source is 1v then this is simply 1/(s^2+s).

Find the inverse of that and you'll be very happy :)
 

WBahn

Joined Mar 31, 2012
30,052
Suppose R = C = 1 then the transfer function from the input voltage to the voltage across the capacitor is 1/ (s+1). So Vc(S) / V(S) = 1/(s+1). Getting back to time-domain: Vc(t) = V(t) * e^-t. What's wrong here? Thank you!
The statement R=C=1 is nonsensical.

That's like saying that the height of a car and the weight of a building are both equal to 1. Totally meaningless.
 

Papabravo

Joined Feb 24, 2006
21,225
C is complex, R isn't.
If C is capacitance measured in farads, then it is a real number. The reactance of C, Xc, with units of ohms is also a real number. The impedance or admitaance of a capacitor C is a complex number. In the ideal case it would have a zero real part. For a real component there will be some equivalent series resistance.
 
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